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Question Number 145979 by iloveisrael last updated on 09/Jul/21

sin^2 x−4cos^3 x−1=2cos^2 x+2cos x−2cos xsin^2 x x=?

$$\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}−\mathrm{4cos}\:^{\mathrm{3}} \mathrm{x}−\mathrm{1}=\mathrm{2cos}\:^{\mathrm{2}} \mathrm{x}+\mathrm{2cos}\:\mathrm{x}−\mathrm{2cos}\:\mathrm{xsin}\:^{\mathrm{2}} \mathrm{x} \\ $$$$\mathrm{x}=? \\ $$

Answered by Olaf_Thorendsen last updated on 10/Jul/21

sin^2 x−4cos^3 x−1 = 2cos^2 x+2cosx−2cosxsin^2 x Let X = cosx 1−X^2 −4X^3 −1 =2X^2 +2X−2X(1−X^2 ) −6X^3 −3X^2 = 0 X^2 (2X+1) = 0 X = cosx = 0 ⇔ x = ±(π/2)+2kπ X = cosx = −(1/2) ⇔ x = ±((2π)/3)+2kπ

$$\mathrm{sin}^{\mathrm{2}} {x}−\mathrm{4cos}^{\mathrm{3}} {x}−\mathrm{1}\:=\:\mathrm{2cos}^{\mathrm{2}} {x}+\mathrm{2cos}{x}−\mathrm{2cos}{x}\mathrm{sin}^{\mathrm{2}} {x} \\ $$$$\mathrm{Let}\:\mathrm{X}\:=\:\mathrm{cos}{x} \\ $$$$\mathrm{1}−\mathrm{X}^{\mathrm{2}} −\mathrm{4X}^{\mathrm{3}} −\mathrm{1}\:=\mathrm{2X}^{\mathrm{2}} +\mathrm{2X}−\mathrm{2X}\left(\mathrm{1}−\mathrm{X}^{\mathrm{2}} \right) \\ $$$$\:−\mathrm{6X}^{\mathrm{3}} −\mathrm{3X}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{X}^{\mathrm{2}} \left(\mathrm{2X}+\mathrm{1}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{X}\:=\:\mathrm{cos}{x}\:=\:\mathrm{0}\:\Leftrightarrow\:{x}\:=\:\pm\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi \\ $$$$\mathrm{X}\:=\:\mathrm{cos}{x}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\:\Leftrightarrow\:{x}\:=\:\pm\frac{\mathrm{2}\pi}{\mathrm{3}}+\mathrm{2}{k}\pi \\ $$

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